Q. 61 - Calculus

Question

Consider the hyperbola with equation $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1.$ Let F be the focus with coordinates $(\sqrt{A^2+B^2}, 0).$ Let $e=\frac{\sqrt{A^2+B^2}}{A}$ and l be the vertical line with equation $x=\frac{A}{e}.$ Show that for any point P on the hyperbola, $\frac{FP}{DP}=e,$ where D is the point on l closest to P.

Step-by-Step Solution

Answer

The given statement is proved.

Step 1. Given Information.

The given hyperbola equation is $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1.$ and the coordinates of focus are $(\sqrt{A^2+B^2}, 0).$

Step 2. Showing.

Let P be the point $\left(x,y\right)$ on the hyperbola.

So, the distance between the points $\left(x,y\right)\left(\sqrt{A^2+B^2},0\right).$

By using the formula for the distance $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}.$

$$\begin{gathered} \mathrm{Here}, x_1=x, y_1=y, x_2=\sqrt{A^2-B^2}, y_2=0 \\ FP=\sqrt{{\left(x-\left(\sqrt{A^2-B^2}\right)\right)}^2+(y-0{)}^2} \\ FP=\sqrt{\left(x^2+A^2-B^2-2x\sqrt{A^2-B^2}\right)+y^2} \\ FP=\sqrt{\left(x^2+A^2-B^2-2x\sqrt{A^2-B^2}\right)-B^2+\frac{B^2{x}^2}{A^2}} \\ \left[\sin ce,\frac{x^2}{A^2}-\frac{y^2}{B^2}=1\Rightarrow y^2=\frac{B^2{x}^2}{A^2}-B^2\right] \\ FP=\sqrt{x^2\left(1+\frac{B^2}{A^2}\right)+A^2-2x\sqrt{A^2+B^2}} \\ FP=\sqrt{\left(\frac{A^2+B^2}{A^2}\right)\left(x^2-2x\sqrt{A^2+B^2}\frac{A^2}{A^2+B^2}+A^2·\frac{A^2}{A^2+B^2}\right)} \\ FP=\sqrt{\frac{A^2+B^2}{A^2}}\sqrt{\left(x^2-2x\sqrt{A^2+B^2}\frac{A^2}{{\left(\sqrt{A^2+B^2}\right)}^2}+A^2·\frac{A^2}{{\left(\sqrt{A^2+B^2}\right)}^2}\right)} \\ FP=\sqrt{\frac{A^2+B^2}{A^2}}\sqrt{\left(x^2-2x\frac{A^2}{\left(\sqrt{A^2+B^2}\right)}+{\left(\frac{A^2}{\left(\sqrt{A^2+B^2}\right)}\right)}^2\right)} \\ FP=\sqrt{\frac{A^2+B^2}{A^2}}\left(x-\frac{A^2}{\sqrt{A^2+B^2}}\right) \\ FP=e\left(x-\frac{A}{e}\right)\left[since\frac{\sqrt{A^2+B^2}}{A}=e\right] \\ \frac{FP}{DP}=\frac{e\left(x-\frac{A}{e}\right)}{x-\frac{A}{e}}\left[\sin ce FP=e\left(x-\frac{A}{e}\right),DP=x-\frac{A}{e}\right] \\ \frac{FP}{DP}=e \end{gathered}$$uncaught exception: Invalid chunk

in file: /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 68
#0 /var/www/html/integration/lib/php/Boot.class.php(769): com_wiris_plugin_impl_HttpImpl_1(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Invalid chunk') #1 /var/www/html/integration/lib/haxe/Http.class.php(532): _hx_lambda->execute('Invalid chunk') #2 /var/www/html/integration/lib/php/Boot.class.php(769): haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Invalid chunk') #3 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(30): _hx_lambda->execute('Invalid chunk') #4 /var/www/html/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Invalid chunk') #5 /var/www/html/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), Object(sys_net_Socket), NULL) #6 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(43): haxe_Http->request(true) #7 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(268): com_wiris_plugin_impl_HttpImpl->request(true) #8 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(307): com_wiris_plugin_impl_RenderImpl->showImage('d5a92cf1d81b51b...', NULL, Object(PhpParamsProvider)) #9 /var/www/html/integration/createimage.php(17): com_wiris_plugin_impl_RenderImpl->createImage('

Step 3. Showing.

By proceeding with the calculation further,

$$\begin{gathered} FP=\sqrt{\left(\frac{A^2+B^2}{A^2}\right)\left(x^2-2x\sqrt{A^2-B^2}\frac{A^2}{A^2+B^2}+A^2·\frac{A^2}{A^2+B^2}\right)} \\ FP=\sqrt{\frac{A^2+B^2}{A^2}}\sqrt{\left(x^2-2x\sqrt{A^2+B^2}\frac{A^2}{{\left(\sqrt{A^2+B^2}\right)}^2}+A^2·\frac{A^2}{{\left(\sqrt{A^2+B^2}\right)}^2}\right)} \\ FP=\sqrt{\frac{A^2+B^2}{A^2}}\sqrt{\left(x^2-2x\frac{A^2}{\left(\sqrt{A^2+B^2}\right)}+{\left(\frac{A^2}{\left(\sqrt{A^2+B^2}\right)}\right)}^2\right)} \\ FP=\sqrt{\frac{A^2+B^2}{A^2}}\left(x-\frac{A^2}{\sqrt{A^2+B^2}}\right) \\ FP=e\left(x-\frac{A}{e}\right)\left[since\frac{\sqrt{A^2+B^2}}{A}=e\right] \\ \mathrm{Now}, \\ \frac{FP}{DP}=\frac{e\left(x-\frac{A}{e}\right)}{x-\frac{A}{e}}\left[\sin ce FP=e\left(x-\frac{A}{e}\right),DP=x-\frac{A}{e}\right] \\ \frac{FP}{DP}=e \end{gathered}$$

Hence proved.